Question:
Initially a gas of diatomic molecules is contained in a cylinder of volume V1 at a pressure P1 and temperature 250 K. Assuming that 25% of the molecules get dissociated causing a change in number of moles. The pressure of the resulting gas at temperature 2000 K, when contained in a volume 2V1 is given by P2 . The ratio \(\frac{P_2}{P_1}\) is ________.
Initially a gas of diatomic molecules is contained in a cylinder of volume V1 at a pressure P1 and temperature 250 K. Assuming that 25% of the molecules get dissociated causing a change in number of moles. The pressure of the resulting gas at temperature 2000 K, when contained in a volume 2V1 is given by P2 . The ratio \(\frac{P_2}{P_1}\) is ________.
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Dissociation of 25% of diatomic molecules raises the number of moles to 1.25 times; then apply PV = nRT before and after.
Updated On: Nov 5, 2026
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Correct Answer: 5
Solution and Explanation
Let the initial number of moles be n. Each dissociated diatomic molecule gives 2 atoms, so with 25% dissociated: \(n'=0.75n+2(0.25n)=1.25n\).
Initially: \(P_1V_1=nR(250)\).
Finally: \(P_2(2V_1)=1.25n\,R(2000)\).
Dividing: \(\frac{2P_2}{P_1}=1.25\times\frac{2000}{250}=1.25\times8=10\).
\(\frac{P_2}{P_1}=5\).
Answer: 5
Initially: \(P_1V_1=nR(250)\).
Finally: \(P_2(2V_1)=1.25n\,R(2000)\).
Dividing: \(\frac{2P_2}{P_1}=1.25\times\frac{2000}{250}=1.25\times8=10\).
\(\frac{P_2}{P_1}=5\).
Answer: 5
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